My Math Forum integral of absolute value of a Fourier transform

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 April 25th, 2017, 05:06 PM #1 Newbie   Joined: Apr 2017 From: New York Posts: 1 Thanks: 0 integral of absolute value of a Fourier transform Hi guys, I am going to calculate the following integral: $$\int_0^{f_c+f_m} |Y(f)|^2\, df$$ where: $$Y(f)=\frac{\pi}{2} \alpha_m \sum_{l=1}^{L} \sqrt{g_l}\left [ e^{-j(\omega \tau_l - \theta_m)} \delta(\omega - \omega_0) + e^{-j(\omega \tau_l + \theta_m)} \delta(\omega + \omega_0) \right ]$$ with $$\omega_0= 2\pi (f_c + f_m), \ \ \alpha_m=constant, \ \ f_c,f_m: frequencies, \ \ \theta_m: initial \ phase$$. Then, the integral weare looking for will get the following form: $$\int_0^{f_c+f_m} |Y(f)|^2 df= \int_o^{f_c + f_m} (\pi \alpha_m)^2 \Big|\sum_{l=1}^L \sqrt{g_l}e^{-j \omega \tau_l} \Big|^2 cos^2[2 \pi (f_c + f_m) + \theta_m]df =\\ (\pi \alpha_m)^2\int_0^{f_c+f_m} \sum_{l=1}^L g_l e^{-2j \omega \tau_l} \Big[cos^2[2 \pi (f_c + f_m) + \theta_m]\Big]df =\\ (\pi \alpha_m)^2 \Big(\sum_{l=1}^L g_l e^{-j2(2\pi) \tau_l}\Big) \Big[cos^2[2 \pi (f_c + f_m) +\theta_m] \Big] \int_0^{f_c+f_m}e^f df$$ Using a delta's Dirac property: ##\delta(\omega - \omega_0)f(\omega)= f(\omega - \omega_0)## (please correct me if it is wrong, because I have doubts about it), I got: $$Y(f)=\frac{\pi}{2} \alpha_m \sum_{l=1}^{L} \sqrt{g_l}\left [ e^{-j[(\omega - \omega_0 )\tau_l - \theta_m)]} + e^{-j[(\omega - \omega_0) \tau_l + \theta_m)]} \right ] =\\ =\frac{\pi}{2} \alpha_m \sum_{l=1}^{L} \sqrt{g_l} e^{-j \omega \tau_l} \left [ e^{j(\omega_0\tau_l + \theta_m)} + e^{-j( \omega_0 \tau_l + \theta_m)]} \right ] =\\ =(\pi \alpha_m) \Big(\sum_{l=1}^{L} \sqrt{g_l} e^{-j \omega \tau_l} \Big) cos [2 \pi (f_c + f_m)\tau_l + \theta_m]$$ So, finally: $$|Y(f)|^2=(\pi \alpha_m)^2 \Big|\sum_{l=1}^L \sqrt{g_l}e^{-j \omega \tau_l} \Big|^2 cos^2[2 \pi (f_c + f_m) + \theta_m]$$. Being ## \int_0^{f_c+f_m}e^f df = e^{f_c+f_m} - 1\approx e^{f_c+f_m} ##, then: $$\int_0^{f_c+f_m} |Y(f)|^2 df= (\pi \alpha_m)^2 \Big(\sum_{l=1}^L g_l e^{-j4 \pi (f_c + f_m) \tau_l}\Big) \Big[cos^2[2 \pi (f_c + f_m) +\theta_m] \Big]$$ My supervisor told me I am supposed to find a solution proportional to: ##\Big|\sum_{l=1}^L \sqrt{g_l}e^{j 2 \pi (f_c + f_m)\tau_l} \Big|^2##. Could you please help me to find the right solution and where the error is? Thank you so much for your help, I would really appreciate that!

 Tags absolute, fourier, integral, transform

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